Ta co 4x2 - 4x = -1
=> 4x2 - 4x + 1 = 0
<=> (2x - 1)2 = 0
=> 2x - 1 = 0
=> 2x = 1
=> x = \(\frac{1}{2}\)
a) \(4x^2-4x=-1\)
\(\Leftrightarrow4x^2-4x+1=0\)
\(\Leftrightarrow\left(2x-1\right)^2=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
b) \(8x^3+12x^2+6x+1=0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right).\left(8x^2+8x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\left(1\right)\\8x^2+8x+2=0\left(2\right)\end{cases}}\)
Giải (1) :
\(x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\)
Giải (2) :
\(8x^2+8x+2=0\)
\(\Leftrightarrow\left(\sqrt{8}x+\sqrt{2}\right)^2=0\)
\(\Leftrightarrow\sqrt{8}x+\sqrt{2}=0\)
\(\Leftrightarrow\sqrt{8}x=-\sqrt{2}\)
\(\Leftrightarrow x=-\frac{\sqrt{2}}{\sqrt{8}}=-\frac{1}{2}\)
Từ (1) ; (2)
\(\Rightarrow x=-\frac{1}{2}\)
